Using the disk method, the volume of the solid is
[tex]\displaystyle \pi \int_3^8 \left(\dfrac1{2x-1}\right)^2 \, dx = \pi \int_3^8 \frac{dx}{(2x-1)^2}[/tex]
Substitute [tex]u = 2x-1[/tex] and [tex]du=2\,dx[/tex].
[tex]\displaystyle \pi \int_3^8 \frac{dx}{(2x-1)^2} = \frac\pi2 \int_5^{15} \frac{du}{u^2} \\\\ ~~~~~~~~ = -\frac\pi2 \frac1u \bigg|_5^{15} \\\\ ~~~~~~~~ = -\frac\pi2 \left(\frac1{15} - \frac15\right) = \boxed{\frac\pi{15}}[/tex]
